Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Mmm so I'm given the problem;

If $f_1, f_2$ are functions with simple poles (degree of 1) at $z_0$, then show $f_1f_2$ has a pole of degree 2 at $z_0$ and provide an expression for $Res(f_1f_2; z_0)$. I.e, the residue of $f_1f_2$ at $z_0$.

This problem is trivial for rational functions, but I'm unsure how to approach this in general. Idea's I've had are to use the Laurent expansions for $f_1, f_2$, but I want to avoid this if possible. The idea is pretty simple, I just don't know of a short, concise way of proving it in general. Other ideas involve showing absolute convergence of the product of the Taylor series expansions of the denominators to rewrite the denominator as a single series where $z_0$ is a pole of degree 2. Not sure if that'd work though..

Thanks for any help!

share|improve this question
    
Why would you want to avoid using Laurent expansions? Anyway, maybe you can try using the derivative/limit formula for the residue (along with the product rule). –  Andrew Nov 19 '12 at 18:28
    
Because, to be honest, I don't quite "get" Laurent expansions the same way I understand Taylor series –  user45814 Nov 19 '12 at 18:33
    
All the more reason to try to use them! It may help in this case to think of a Laurent series for $f_i$ as $f_i(z)=\dfrac{a_{i,-1}}{z-z_0}+g_i(z)$ where $g_i(z)$ is holomorphic near $z_0$ and $i=1,2.$ Multiplying the series for $f_1$ with $f_2$ behaves exactly as you'd hope. –  Andrew Nov 19 '12 at 18:39

1 Answer 1

up vote 1 down vote accepted

I know you said you wanted to avoid Laurent series, but it really isn't too painful. I'll expand on Andrew's comment for the possibility that seeing more details will help you out.

Since $f_1$ and $f_2$ have simple poles at $z=z_0$ they can be written as Laurent series,

$$ f_1(z) = \sum_{k=-1}^{\infty} a_k (z-z_0)^k, $$ $$ f_2(z) = \sum_{k=-1}^{\infty} b_k (z-z_0)^k, $$

which converge in some annulus $0 < |z-z_0| < R$. Multiplying these series term-by-term gives another Laurent series which converges in this annulus,

$$ f_1(z)f_2(z) = \sum_{k=-2}^{\infty} \left(\sum_{p+q = k} a_p b_q \right) (z-z_0)^k, $$

which shows that $f_1f_2$ has a pole of order 2 at $z=z_0$. This pole has residue

$$ \sum_{p+q = -1} a_p b_q = a_{-1}b_0 + a_0b_{-1}, $$

which is the coefficient of $(z-z_0)^{-1}$ in this Laurent series.

(Note that the inner sum is the Cauchy product of the relevant coefficients of the individual Laurent series.)

If you'd like, these coefficients $a_{-1},a_0,b_{-1},b_0$ can be written down explicitly in terms of $f_1$ and $f_2$. For example, it follows from the Laurent expansion of $f_1$ that

$$ a_{-1} = \lim_{z \to z_0} f_1(z)(z-z_0) $$

and

$$ a_0 = \lim_{z \to z_0} \left(f_1(z) - \frac{a_{-1}}{z-z_0}\right). $$

share|improve this answer
    
very thorough answer, appreciated! I'm going to need some time to digest this a little though as my understanding of Laurent series is shaky. One question though, should I explicitly indicate each of the Laurent series $f_1(z), f_2(z)$ converges absolutely in order to use the "term-by-term" multiplication of infinite series? –  user45814 Nov 19 '12 at 19:04
    
Yes, that's definitely why you can multiply like that. Power series (both Taylor and Laurent) are really nice in that respect; they converge absolutely within their (open) domains of convergence. –  Antonio Vargas Nov 19 '12 at 19:15
    
Yessssss, thats awesome. Thanks! –  user45814 Nov 19 '12 at 22:36
    
@user45814, You're very welcome :) –  Antonio Vargas Nov 20 '12 at 2:46

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.